Divide using synthetic division $$ ( 3x^{3}+x^{2}-6x+2 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&3&1&-6&2\\& & 3& 4& \color{black}{-2} \\ \hline &\color{blue}{3}&\color{blue}{4}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+x^{2}-6x+2 }{ x-1 } = \color{blue}{3x^{2}+4x-2} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&1&-6&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 3 }&1&-6&2\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&1&-6&2\\& & \color{blue}{3} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 3 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}1&3&\color{orangered}{ 1 }&-6&2\\& & \color{orangered}{3} & & \\ \hline &3&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&1&-6&2\\& & 3& \color{blue}{4} & \\ \hline &3&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 4 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}1&3&1&\color{orangered}{ -6 }&2\\& & 3& \color{orangered}{4} & \\ \hline &3&4&\color{orangered}{-2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&3&1&-6&2\\& & 3& 4& \color{blue}{-2} \\ \hline &3&4&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}1&3&1&-6&\color{orangered}{ 2 }\\& & 3& 4& \color{orangered}{-2} \\ \hline &\color{blue}{3}&\color{blue}{4}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+4x-2 } $ with a remainder of $ \color{red}{ 0 } $.